设AB的中点为O(x,y);A(x1,y1),B(x2,y2)
直线过抛物线y^2=4x得焦点,而焦点F(1,0)
设直线的方程为:y=k(x-1)
.........................(1)
将(1)^2代入抛物线方程中可得:
k^2(x-1)^2=4x
=>k^2x^2-(2k^2+4)x+k^2=0
x1+x2=(2k^2+4)/k^2
y1+y2=k(x1+x2-2)=4/k
..............................(2)
又
x=(x1+x2)/2=(k^2+2)/k^2=(2+(2/k^2)).................(3)
y=(y1+y2)/2=2/k
=>2/k^2=y^2/2.........................(4)
将(4)代入(3)可得
x=(2+(y^2/2))
=>y^2=2x-4
所以
AB的中心轨迹方程为:y^2=2x-4
5条
二哥你也太囧了
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